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pychu [463]
3 years ago
5

A woman 5.5 ft walks at a rate of 6 ft/sec towards a street light that is 22 ft above the ground. At what rate is the length of

her shadow changing when she is 15 ft from the base of the light?
Physics
2 answers:
Soloha48 [4]3 years ago
7 0

Answer:

The length of her shadow is changing at the rate  -2 m/s

Explanation:

Let the height oh the street light, h = 22 ft

Let the height of the woman, w = 5.5 ft

Horizontal distance to the street light = l

length of shadow = x

h/w = (l + x)/x

22/5.5 =  (l + x)/x

4x = l + x

3x = l

x = 1/3 l

taking the derivative with respect to t of both sides

dx/dt = 1/3 dl/dt

dl/dt = -6 ft/sec ( since the woman is walking towards the street light, the value of l is decreasing with time)

dx/dt = 1/3 * (-6)

dx/dt = -2 m/s

sveticcg [70]3 years ago
5 0

Answer:

Explanation:

By applying similar triangles rule,

22/5.5 = (x + s)/s

s = 5.5/22 × (x + s)

ds/dt = 5.5/22 × (dx/dt + ds/dt)

16.5/22 × ds/dt = 5.5/22 × dx/dt

16.5/22 × ds/dt = 5.5/22 × -6

ds/st = -2 ft/s.

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